15 THE LEBESGUE-NIKODYM THEOREM 195 On the other hand, if p 2> p for all // e H, then in particular p g v0 , by the definition of v0 ; hence v0 is the least upper bound v, and we have v(/) = sup n(f) He H for all/£ 0 belonging to JfR(X). (13.15.5) (Lebesgue-Nikodym Theorem) Let ^, v be two positive measures on X. Then the following conditions are equivalent: (a) v is a measure with base /j,. (b) Every ^-negligible set is v-negligible. (b') Every compact ^-negligible set is v-negligible. (c) For every function /^ 0 which is both /n-integrable and v-integrable, and for every real number e > 0, there exists 5 > 0 such that the relations 0 £ A £/a/?rf J*/z ^u g 5 7/H/>/y J*/z dv £ s. (c') /for every compact subset K 0/*X a«rf euery /*£#/ number s > 0, //z^re tfjcto <5 > 0 ^wc/z //?#/ £/z£ relations A c K and /^*(A) ^ (5 /w/?/y v*(A) ^ e ("absolute continuity" of v with respect to fj). (d) v = sup (inf(v, The fact that (a) implies (b) follows immediately from (13.14.1) and (13.6.3). Clearly (b) implies (br). Conversely, suppose that (b') is satisfied, and let N be a /^-negligible set. Since X is a denumerable union of compact sets Kn, it is enough to show that each of the sets N n Krt is v-negligible, and we may therefore assume that N is relatively compact. Since N then has a compact neighborhood (3.18.2), it follows from (13.7.9) and (13.8.7, (i)) that we may restrict ourselves to the case where N is a denumerable intersection of relatively compact open sets; but in this case N is v-integrable ((13.7.7) and (13.8.7, (i)), hence is the union of a sequence (Hw) of compact sets and a v-negligible set P. Since by hypothesis v(HJ = 0 for all n, it follows that N is v-negligible. To prove that (b) implies (d) and that (d) implies (a), we remark that we may express n and v in the form /j, = u - X and v = v -1, where A is a positive measure and u, v are finite, ^0 and locally 1-integrable ((13.15.2) and (13.15.3)). If we put v' = sup(inf(y, nu)) and v" = v — v', then we have v' g 0, v" S> 0 and sup(inf(v, WJM)) = v' -1 (13.15.3). Hence to show that (b) n implies (d) it is enough to show that (b) implies that v" -1 = 0. Now the set A of points x e X at which v"(x) > 0 is contained in the set of points jc at whiche